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- 1r2
- 1r
- 1r13/2
- 1r13/5

1r13/2

1r

1r13/5

1r2

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Solution

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∫E.dS=qencϵ0

∴E×2πr×l=λ×lϵ0

∴E=λ2πrϵ0

∴E×2πr×l=λ×lϵ0

∴E=λ2πrϵ0

Here λ is the linear charge density on the wire.

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